Anderson Video - Induced Current in Wire Loop

Professor Anderson
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Hello class, Professor Anderson here let's take a look at a good induction problem. Let's say we have a square metal loop and we're going to pull it into a region of B field. Now a square metal loop will have some internal resistance to it, and let's call that internal resistance R. And let's see if we can figure out what happens to the current in the loop as it starts to cross this intersection between B field and no B field Alright so how do we think about this? Well this is, of course, induction, and what do we know about induction? What we know is that the EMF epsilon is negative d(phi)/dt, where this thing phi is the flux, it's the B field flux. Alright. B field flux is just simply the strength of the B field times the area. So if we put that into this equation, what do we get? We get negative d/dt of B times A. But B isn't changing, right? B is just the strength of the B field that's given, it doesn't change as a function of time so we get negative B times dA/dt so this area is key. We've got to figure out how much area is intercepting B field. Alright, maybe we can do that. If I think about this loop coming across this plane, what can I say? Well, let's extend our B field plane down a little bit more, and let's put some loops down here indicating this loop moving into the B field. Alright so right here it starts to cross and then at a later time it's further in, and so forth. So if we can identify the key area which is this area right there then maybe we can figure out how that area is changing in time. Alright how do we do that, how do we figure out what that area is? Well, let's identify a parameter for it, and I'm going to take this little triangle here, that shaded area, and I'm gonna blow it up right over here. Okay, so here is that triangular region. Let's call this side x. Okay, as it moves in, it moves in a distance x. And now, by symmetry, you can see, well if that's x I know this thing is a 45-45-90 triangle. This side right here has to be X. Alright so what is the area of this whole thing? Well the area of this little triangle up here is going to be 1/2 the base times the height. Which is 1/2 x times x so I get an x squared. But the area of this little triangle down here is also 1/2 x squared. And so look what I get, I just get the area of this whole thing is X squared. Alright that looks pretty cool. What if I do a derivative of that area? So I want to take the time derivative of x squared. What do I get? Well I pull down a two, I get an x but I have to remember that x is a function of time and so I get dx/dt. But I know what all those things are. If this is the distance x, I know that x is just equal to how fast it's moving times time. Alright, so what does dA/dt become? dA/dt becomes two times X which we said is vt times dx/dt which is just v. And so what does this thing become? It becomes two v squared times t. Alright that looks pretty cool. Maybe we can put this together with this to figure out something about the current. Alright, as the loop starts to come in and the B field flux changes it's going to generate current in there, don't worry about which direction the current is. If you're really concerned about that check out the video on Lenz's law. But let's see if we can figure out what the amplitude is of that current. Okay induction EMF, right. EMF Epsilon that's just like a voltage. Okay, so if we have a voltage and we have a resistance then we have to have a current. So I'm gonna leave that there for a second. We know that the EMF is equal to I times R. So what is I? It is epsilon over R. But now we know what epsilon is. It's negative B dA/dt and we're going to divide by R and this is our dA/dt. So we get negative B times two times v squared times t and we're going to divide by R. Okay, this is the current as a function of time. Again the negative sign just tells you about direction, whether it's going this way or this way, and Lenz's law will tell you what that is. Okay that's our current, how do we figure out how to map this thing out as a function of time? Well, first off we need to think about how the area is changing as a function of time, right? It''s starting to ramp up quite quickly until it gets to that point and then it gets diminishing returns, right? It starts to change less rapidly on the backside. And so it turns out the maximum current is going to be right at the halfway point, right when this square loop gets there. Okay, when it gets to the halfway point, that's when it is a maximum and so we need to figure out when that happens in time. Alright, to do that, let's look at this triangle again. If I think about this triangle now this side is l. What is this side? Well, this is a one one root two triangle so this side down here must be l over root two. So how long does it take for this thing to advance that distance? Alright that's not too bad. It takes a time equal to how far you've gone, l over root two divided by how fast you do it: v. This is the time to reach the peak of the current. Okay, and so now you can figure out what the peak current is. I peak is going to be this thing, negative B times two times v squared times t peak all over R. And let's not worry about the sign, so we're just gonna say the magnitude of I, and then I'm going to take off this negative sign right there. And we'll say the magnitude there and we'll take off that negative sign right there. Okay, this is how you calculate what the peak current is and now you can plot this thing out as a function of time. Here is I, here is t and what does it look like? Well, it goes up until it hits I peak and that happens at a time t peak, and then it goes down linearly until it comes back down to zero in the same amount of time. Okay, and then it's a zero from there on out. So that's what your curve looks like as a function of time. Okay, you know it has to be symmetric about t peak because if you ran the whole thing in reverse it should go up and then down just like this behavior. Alright hopefully that's clear. You guys can plug in some numbers and try it out and see what you get. Cheers.
Hello class, Professor Anderson here let's take a look at a good induction problem. Let's say we have a square metal loop and we're going to pull it into a region of B field. Now a square metal loop will have some internal resistance to it, and let's call that internal resistance R. And let's see if we can figure out what happens to the current in the loop as it starts to cross this intersection between B field and no B field Alright so how do we think about this? Well this is, of course, induction, and what do we know about induction? What we know is that the EMF epsilon is negative d(phi)/dt, where this thing phi is the flux, it's the B field flux. Alright. B field flux is just simply the strength of the B field times the area. So if we put that into this equation, what do we get? We get negative d/dt of B times A. But B isn't changing, right? B is just the strength of the B field that's given, it doesn't change as a function of time so we get negative B times dA/dt so this area is key. We've got to figure out how much area is intercepting B field. Alright, maybe we can do that. If I think about this loop coming across this plane, what can I say? Well, let's extend our B field plane down a little bit more, and let's put some loops down here indicating this loop moving into the B field. Alright so right here it starts to cross and then at a later time it's further in, and so forth. So if we can identify the key area which is this area right there then maybe we can figure out how that area is changing in time. Alright how do we do that, how do we figure out what that area is? Well, let's identify a parameter for it, and I'm going to take this little triangle here, that shaded area, and I'm gonna blow it up right over here. Okay, so here is that triangular region. Let's call this side x. Okay, as it moves in, it moves in a distance x. And now, by symmetry, you can see, well if that's x I know this thing is a 45-45-90 triangle. This side right here has to be X. Alright so what is the area of this whole thing? Well the area of this little triangle up here is going to be 1/2 the base times the height. Which is 1/2 x times x so I get an x squared. But the area of this little triangle down here is also 1/2 x squared. And so look what I get, I just get the area of this whole thing is X squared. Alright that looks pretty cool. What if I do a derivative of that area? So I want to take the time derivative of x squared. What do I get? Well I pull down a two, I get an x but I have to remember that x is a function of time and so I get dx/dt. But I know what all those things are. If this is the distance x, I know that x is just equal to how fast it's moving times time. Alright, so what does dA/dt become? dA/dt becomes two times X which we said is vt times dx/dt which is just v. And so what does this thing become? It becomes two v squared times t. Alright that looks pretty cool. Maybe we can put this together with this to figure out something about the current. Alright, as the loop starts to come in and the B field flux changes it's going to generate current in there, don't worry about which direction the current is. If you're really concerned about that check out the video on Lenz's law. But let's see if we can figure out what the amplitude is of that current. Okay induction EMF, right. EMF Epsilon that's just like a voltage. Okay, so if we have a voltage and we have a resistance then we have to have a current. So I'm gonna leave that there for a second. We know that the EMF is equal to I times R. So what is I? It is epsilon over R. But now we know what epsilon is. It's negative B dA/dt and we're going to divide by R and this is our dA/dt. So we get negative B times two times v squared times t and we're going to divide by R. Okay, this is the current as a function of time. Again the negative sign just tells you about direction, whether it's going this way or this way, and Lenz's law will tell you what that is. Okay that's our current, how do we figure out how to map this thing out as a function of time? Well, first off we need to think about how the area is changing as a function of time, right? It''s starting to ramp up quite quickly until it gets to that point and then it gets diminishing returns, right? It starts to change less rapidly on the backside. And so it turns out the maximum current is going to be right at the halfway point, right when this square loop gets there. Okay, when it gets to the halfway point, that's when it is a maximum and so we need to figure out when that happens in time. Alright, to do that, let's look at this triangle again. If I think about this triangle now this side is l. What is this side? Well, this is a one one root two triangle so this side down here must be l over root two. So how long does it take for this thing to advance that distance? Alright that's not too bad. It takes a time equal to how far you've gone, l over root two divided by how fast you do it: v. This is the time to reach the peak of the current. Okay, and so now you can figure out what the peak current is. I peak is going to be this thing, negative B times two times v squared times t peak all over R. And let's not worry about the sign, so we're just gonna say the magnitude of I, and then I'm going to take off this negative sign right there. And we'll say the magnitude there and we'll take off that negative sign right there. Okay, this is how you calculate what the peak current is and now you can plot this thing out as a function of time. Here is I, here is t and what does it look like? Well, it goes up until it hits I peak and that happens at a time t peak, and then it goes down linearly until it comes back down to zero in the same amount of time. Okay, and then it's a zero from there on out. So that's what your curve looks like as a function of time. Okay, you know it has to be symmetric about t peak because if you ran the whole thing in reverse it should go up and then down just like this behavior. Alright hopefully that's clear. You guys can plug in some numbers and try it out and see what you get. Cheers.